Question: Solve for $c$. $\dfrac54=-4c+\dfrac14 $
Solution: Let's subtract and then divide to get $c$ by itself. $\begin{aligned} \dfrac54&=-4c+\dfrac14 \\ \\ \dfrac54{-\dfrac14} &=-4c+\dfrac14 {-\dfrac14}~~~~~~{\text{subtract }\dfrac14} \text{ from each side}\\ \\ \dfrac54{-\dfrac14}&=-4c+\cancel{ \dfrac14} {{-}\cancel{{\dfrac14}}}\\ \\ \dfrac54{-\dfrac14}&=-4c \end{aligned}$ $\begin{aligned}1&= -4c \\ \\ \dfrac{1}{{-4}} &= \dfrac{-4c}{{-4}} ~~~~~~~\text{divide each side by } {-4} \text{ to get } c \text{ by itself }\\ \\ \dfrac{1}{{-4}} &= \dfrac{\cancel{-4}c}{\cancel{{-4}}} \\ \\ \dfrac{1}{{-4}}&= c \end{aligned}$ The answer: $c={-\dfrac14}$ Let's check to make sure. $\begin{aligned} \dfrac54&=-4c+\dfrac14 \\\\ \dfrac54&\stackrel{?}{=} -4\left({-\dfrac14}\right)+\dfrac14 \\\\ \dfrac54&\stackrel{?}{=} 1+\dfrac14 \\\\ \dfrac54 &= \dfrac54 ~~~~~~~~~~\text{Yes!} \end{aligned}$